How many ways can 1995 be factored as a product of two two-digit numbers? (Two factorizations of the form $a\cdot b$ and $b\cdot a$ are considered the same).
Solution: $1995=5\cdot399=3\cdot5\cdot133=3\cdot5\cdot7\cdot19$. Since $3\cdot5\cdot7=105$ has three digits, in any expression of $1995$ as the product of two two-digit numbers, $19$ must be a proper factor of one of them. $19\cdot3=57$ and $19\cdot5=95$ are two-digit numbers that are divisible by $19$ and divide $1995$, but $19\cdot7=133$ and $19\cdot3\cdot5=285$ are three digits, so the only possible expressions of $1995$ as the product of two two-digit numbers are $57\cdot35$ and $95\cdot21$. So, there are $\boxed{2}$ such factorizations.